Divide using synthetic division $$ ( x^{4}-5x^{3}+2x^{2}-7 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&1&-5&2&0&-7\\& & -1& 6& -8& \color{black}{8} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{8}&\color{blue}{-8}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ x^{4}-5x^{3}+2x^{2}-7 }{ x+1 } = \color{blue}{x^{3}-6x^{2}+8x-8} ~+~ \frac{ \color{red}{ 1 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-5&2&0&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 1 }&-5&2&0&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-5&2&0&-7\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-1&1&\color{orangered}{ -5 }&2&0&-7\\& & \color{orangered}{-1} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-5&2&0&-7\\& & -1& \color{blue}{6} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 6 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-1&1&-5&\color{orangered}{ 2 }&0&-7\\& & -1& \color{orangered}{6} & & \\ \hline &1&-6&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 8 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-5&2&0&-7\\& & -1& 6& \color{blue}{-8} & \\ \hline &1&-6&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-1&1&-5&2&\color{orangered}{ 0 }&-7\\& & -1& 6& \color{orangered}{-8} & \\ \hline &1&-6&8&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-5&2&0&-7\\& & -1& 6& -8& \color{blue}{8} \\ \hline &1&-6&8&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 8 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-1&1&-5&2&0&\color{orangered}{ -7 }\\& & -1& 6& -8& \color{orangered}{8} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{8}&\color{blue}{-8}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}+8x-8 } $ with a remainder of $ \color{red}{ 1 } $.